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Some Remarks on Bielliptic and Trigonal Curves

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Some Remarks on Bielliptic and Trigonal Curves Some remarks on bielliptic and trigonal curves BY Andreas Schweizer1 Department of Mathematics, Korea Advanced Institute of Science and Technology (KAIST), Daejeon 305-701 South Korea e-mail: [email protected] Abstract We prove some results on algebraic curves X of genus g 2 in characteristic ≥ 0. For example: Assume that X has an automorphism σ of prime order p 5. ≥ If σ has no fixed points, then X cannot be trigonal. On the other hand, if σ has fixed points, then X is bielliptic only if it belongs to one of three extremal types of curves of small genus. Mathematics Subject Classification (2010): primary 14H45, secondary 14H37, 30F10 Key words: bielliptic curve; trigonal curve; automorphism; smooth covering; fixed point; cubic point 1. Introduction and basic facts Unless said otherwise, in this paper curve means a connected, smooth, projective, algebraic curve over the complex numbers. Equivalently, one can think of a compact Riemann surface. In [M], using the uniformization of compact Riemann surfaces of genus g 2 ≥ by Fuchsian groups, Maclachlan proved, among other results, the following theorem. arXiv:1512.07963v1 [math.AG] 25 Dec 2015 Theorem 1.1. (Maclachlan) [M, Theorem 2] Let X be a hyperelliptic curve and G a subgroup of Aut(X) such that the covering X X/G is totally unramified. Then G divides 4 and G has exponent 2. → | | We first give a short and elementary proof of this theorem. Then we work out similar and related results for other classes of curves, notably for bielliptic and trig- onal curves. Proof. Denote the genus of X by g. The hyperelliptic involution τ lies in the center of Aut(X). So G acts on the 2g +2 fixed points of τ, and as the action of G is without fixed points, G divides 2g + 2. On the other hand, since the covering is | | 1The author was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MSIP) (ASARC, NRF-2007-0056093). 1 unramified, the Hurwitz formula implies that G divides 2g 2. Subtraction already | | − shows that G divides 4. With an| eye| towards generalization, we prove the second claim independently of the first by counting the ramified points on X of the covering X X/ τ,G . → h i The subcover X/ G X/ τ,G is the hyperelliptic one and has 2h+2 ramified points, where h is theh genusi→ ofhX/ Gi . They all split completely in X, which gives G (2h +2)=2g +2+4 G 4 ramifiedh i points, each with ramification index 2. | | | | − Taking the 2g 2 ramified points of X X/ τ into account, there must still be 2 G 2 ramified− points in X/ τ X/→τ,G h, eachi with ramification index 2 | | − h i → h i and splitting into 2 points in X. Note that the covering X/ τ X/ τ,G is Galois h i→ h i with group Ge induced by (and isomorphic to) G, but not necessarily fixed-point- free. The stabilizer in G of each of these 2 G 2 points on X/ τ is an involution. e | | − h i Since X/ τ has genus 0, each such involution has exactly 2 fixed points. So Ge (and hence G)h musti have at least G 1 involutions, i.e. every non-trivial element is an involution. | | − Maclachlan’s theorem says that hyperelliptic curves cannot have automorphisms of certain types, for example fixed-point-free of prime order p 3. In the sequel we want to generalize this to other types of curves. ≥ On the other hand, it is clear that the curve Y = X/G in Theorem 1.1 must also be hyperelliptic. So another interpretation is that certain coverings X of a hyperelliptic curve Y cannot be hyperelliptic. In this form we also want to generalize it, keeping the curve Y hyperelliptic, but taking other types of curves X. One generalization of a hyperelliptic involution is the notion of a b-hyperelliptic involution, that is, an involution such that the genus of the quotient curve is b. We concentrate on the case b = 1; then the involution and the curve are called bielliptic. Some of the literature also uses the expression elliptic-hyperelliptic for this. Another generalization of X being hyperelliptic is the existence of an n-gonal map. This is a surjective morphism of degree n from X to a projective line. Note that for n > 2 such a morphism is not necessarily Galois, i.e. does not necessarily come from an automorphism of X. This makes it much more difficult to establish or exclude the existence of such a map. We are mainly interested in the case n = 3; then the map and the curve are called trigonal. Throughout the paper we will freely use the following easy consequences of the Castelnuovo inequality: The hyperelliptic involution is unique. • On a bielliptic curve of genus g 6 the bielliptic involution is unique. • ≥ A curve of genus g 5 has at most one trigonal map. • ≥ A hyperelliptic curve of genus g 3 cannot be trigonal. • ≥ 2 A hyperelliptic curve of genus g 4 cannot be bielliptic. • ≥ Finally, we’ll need the following results. Theorem 1.2. Let Y be a hyperelliptic curve of genus h, and let X Y be a → smooth, cyclic covering of degree n. (a) (Bujalance) [B, Theorem] If n =2, then X is b-hyperelliptic for some b with with b h−1 . ≤ ⌊ 2 ⌋ In particular, a smooth degree 2 cover of a genus 2 curve is a genus 3 curve that is hyperelliptic (and bielliptic). (b) (Accola) [A, Lemma 2] If n is odd, then the hyperelliptic involution of Y lifts (n−1)(h−1) to n different involutions on X that are all b-hyperelliptic where b = 2 . Together with the automorphism of order n from Gal(X/Y ) they generate a dihedral group Dn of order 2n. In particular, a smooth degree 3 Galois cover of a genus 2 curve is a genus 4 curve that has at least 3 bielliptic involutions. 2. Bielliptic curves We start with a quick corollary to the known results from the previous section. Corollary 2.1. Let X be a curve of genus 5. If X has an automorphism σ of order 4 such that σ has no fixed points, then X is bielliptic. Proof. As X/ σ2 is an unramified degree 2 cover of X/ σ , the genus of X/ σ2 h i h i h i can only be 1 (if σ2 has fixed points) or 3. In the first case we are done. In the second case we get a chain X X/ σ2 X/ σ of smooth coverings of degree 2 with genera 5, 3, and 2. By Theorem→ h i 1.2 → (a) theh i genus 3 curve is hy- perelliptic, and the genus 5 curve is hyperelliptic or bielliptic. But it cannot be hyperelliptic by Theorem 1.1. Ultimately Theorem 1.2 and Corollary 2.1 are based on lifting the hyperelliptic involution to an unramified cover. Our next result requires lifting the hyperelliptic involution to a ramified cover, which is a much more tricky problem. The paper [CT] discusses conditions under which this is possible. Compared to our previous statements we change the names of the curves and their automorphisms so that they fit precisely with those in the somewhat technical conditions in [CT]. Proposition 2.2. Let Y be a non-hyperelliptic curve of genus 4. If Y has an automorphism τ of order 4, then Y is bielliptic. 3 Proof. Since Y is non-hyperelliptic, the curve X = Y/ τ 2 can only have genus 1 h i or 2. In the first case we are done. In the second case τ 2 has exactly 2 fixed points P and Q on Y . From the Hurwitz formula we see that Y/ τ has genus 1 and τ also fixes P and Q. So τ, the involution h i induced by τ on X is different from σ, the hyperelliptic involutione of X. As σ commutes with τ, it acts on its fixed points Pe and Qe, the images of P and Q on X. Moreover, differente involutions must have disjoint fixed points. Hence σ(Pe) = Qe. Since the Weierstrass points of X are exactly the fixed points of σ, we see that Pe and Qe = σ(Pe), the only two points on X that ramify in Y , are not Weierstrass points of X. So our constellation satifies the conditions of [CT, Theorem 2.1], which allows us to conclude that σ lifts to an automorphism of Y . Actually, σ cannot lift to an automorphism ψ of order 4, because then we would have ψ2 = τ 2 and as before X/ σ = Y/ ψ would have genus 1. So σ lifts to an involution ψ on Y that commutesh i with τ 2h. Theni the other two intermediate curves between Y and the genus zero curve Y/ τ 2, ψ have genera 2 and 0 (which would h i contradict non-hyperellipticity of Y ) or 1 and 1, which implies that in this case Y actually has at least 2 bielliptic involutions. Here is an analogue of Maclachlan’s result for bielliptic curves. Proposition 2.3. Let Y be a hyperelliptic curve of genus at least 4 and let X Y be a smooth Galois cover with Galois group G such that X is bielliptic. Then → (a) At least half of the elements of G are involutions. (b) If G is abelian, then G has exponent 2 and G divides 8.
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